∖S. Thus T consists of all integers larger than or equal to a
which are not in S. The theorem will be proved if T = ∅. If T≠∅ then by the well ordering
principle, there would have to exist a smallest element of T, denoted as b. It must be the case
that b > a since by definition, a

∖ S = T which contradicts the choice of b as the smallest element of
T. (b − 1 is smaller.) Since a contradiction is obtained by assuming T≠∅, it must
be the case that T = ∅ and this says that everything in [a,∞) ∩ ℤ is also in S.■

Mathematical induction is a very useful device for proving theorems about the
integers.

Example 2.7.4Prove by induction that∑k=1nk2 =

n(n+1)(2n+1)
-----6----

.

By inspection, if n = 1 then the formula is true. The sum yields 1 and so does the formula
on the right. Suppose this formula is valid for some n ≥ 1 where n is an integer.
Then

∑k=1n+1k2

= ∑k=1nk2 +

(n+ 1)

2

=

n(n-+-1)-(2n-+-1)
6

+

(n +1)

2.

The step going from the first to the second line is based on the assumption that the
formula is true for n. This is called the induction hypothesis. Now simplify the expression in
the second line,

n(n+-1)(2n-+-1)- 2
6 + (n+ 1) .

This equals

( )
(n + 1) n(2n+-1)-+(n + 1)
6

and

n(2n+-1)-
6

+

(n+ 1)

=

2
6(n+-1)+-2n--+n-
6

=

(n-+-2)(2n+-3)-
6

Therefore,

∑k=1n+1k2

=

(n + 1)(n+ 2)(2n+ 3)
---------6----------

=

(n + 1)((n + 1)+ 1)(2(n+ 1)+ 1)
--------------6---------------

,

showing the formula holds for n + 1 whenever it holds for n. This proves the formula by
mathematical induction.

Example 2.7.5Show that for all n ∈ ℕ,

12

⋅

34

⋅⋅⋅

2n2−n1

<

√-1--
2n+1

.

If n = 1 this reduces to the statement that

12

<

√13-

which is obviously true. Suppose then
that the inequality holds for n. Then

and this is clearly true which may
be seen from expanding both sides. This proves the inequality.

Lets review the process just used. If S is the set of integers at least as large as 1 for which
the formula holds, the first step was to show 1 ∈ S and then that whenever n ∈ S, it
follows n + 1 ∈ S. Therefore, by the principle of mathematical induction, S contains
[1,∞) ∩ ℤ, all positive integers. In doing an inductive proof of this sort, the set, S is
normally not mentioned. One just verifies the steps above. First show the thing is
true for some a ∈ ℤ and then verify that whenever it is true for m it follows it is
also true for m + 1. When this has been done, the theorem has been proved for all
m ≥ a.

Proof: Let x be the smallest positive integer. Not surprisingly, x = 1 but this can be
proved. If x < 1 then x2< x contradicting the assertion that x is the smallest natural
number. Therefore, 1 is the smallest natural number. This shows there is no integer, y,
satisfying x < y < x + 1 since otherwise, you could subtract x and conclude 0 < y −x < 1 for
some integer y − x.

You probably saw the process of division in elementary school. Even though you saw it at
a young age it is very profound and quite difficult to understand. Suppose you want to do the
following problem

7292

. What did you do? You likely did a process of long division which gave
the following result.

79
22 = 3 with remainder 13.

This meant

79 = 3 (22) +13.

You were given two numbers, 79 and 22 and you wrote the first as some multiple of the
second added to a third number which was smaller than the second number. Can this always
be done? The answer is in the next theorem and depends here on the Archimedean property
of the real numbers.

Theorem 2.7.11Suppose 0 < a and let b ≥ 0. Then there exists a uniqueinteger p and real number r such that 0 ≤ r < a and b = pa + r.

Proof: Let S ≡

{n ∈ ℕ : an > b}

. By the Archimedean property this set is nonempty. Let
p + 1 be the smallest element of S. Then pa ≤ b because p + 1 is the smallest in S.
Therefore,

r ≡ b− pa ≥ 0.

If r ≥ a then b − pa ≥ a and so b ≥

(p + 1)

a contradicting p + 1 ∈ S. Therefore, r < a as
desired.

To verify uniqueness of p and r, suppose pi and ri, i = 1,2, both work and r2> r1. Then a
little algebra shows

p1 − p2 = r2-−-r1-∈ (0,1).
a

Thus p1− p2 is an integer between 0 and 1, contradicting Theorem 2.7.8. The case that
r1> r2 cannot occur either by similar reasoning. Thus r1 = r2 and it follows that p1 = p2.■

This theorem is called the Euclidean algorithm when a and b are integers. In this case, you
would have r is an integer because it equals an integer.